A sample of 0.1g of water at 100 ^ o C and normal pressure (1.013 ×...

The values are given in the above question, $∆ Q = 54cal = 54 \times 4.18joule = 225.72joule$
$∆ W = P\left[{V}_{steam} - {V}_{water}\right]\left[Forwater0.1gram = 0.1\text{c}\text{c}\right]$
$∆ W = 1.013 \times {10}^{5}\left[167.1 \times {10}^{- 6} - 0.1 \times {10}^{- 6}\right]joule$
$⇒ ∆ W = 1.013 \times 167 \times {10}^{- 1} = 16.917joule$
. By first law of thermodynamics, the change in internal energy,
$⇒ ∆ U = ∆ Q - ∆ W = 225 .72 - 16 .917 = 208 .8 J$